Ruby类，用于简单的版本号自然顺序排序

Posted 2021-02-26

Probably not the most elegant solution, but it works. Needed this for a capistrano deploy task which shows the most recent tagged releases in my repository. Very bare bones, and needs tweaking if your versions are not in X.X.X.X format.

class Version
  include Comparable
 
  attr_reader :major, :feature_group, :feature, :bugfix
 
  def initialize(version="")
    v = version.split(".")
    @major = v[0].to_i
    @feature_group = v[1].to_i
    @feature = v[2].to_i
    @bugfix = v[3].to_i
  end
 
  def <=>(other)
    return @major <=> other.major if ((@major <=> other.major) != 0)
    return @feature_group <=> other.feature_group if ((@feature_group <=> other.feature_group) != 0)
    return @feature <=> other.feature if ((@feature <=> other.feature) != 0)
    return @bugfix <=> other.bugfix
  end
 
  def self.sort
    self.sort!{|a,b| a <=> b}
  end
 
  def to_s
    @major.to_s + "." + @feature_group.to_s + "." + @feature.to_s + "." + @bugfix.to_s
  end
end
 
## Example Usage:
##  list = []
##  ["1.2.2.10","1.2.2.1","2.1.10.2","2.3.4.1"].each {|v| list.push(Version.new(v)) }
##  list.sort.each{|v| pp v.to_s }
##
###"1.2.2.1"
###"1.2.2.10"
###"2.1.10.2"
###"2.3.4.1"